Alpha Gemini Solutions
1. C
2. C
3. B
4. C
5. A
Mu Alpha Theta National Convention 2012
6. E
7. D
8. A
9. B
10. D
11. A
12. D
13. C
14. B
15. B
16. B
17. D
18. C
19. D
20. C
21. B
22. A
23. A
24. B
25. D
1. C. By the sum/difference formulas, this is sin  63  18   sin  45  
26. E
27. C
28. A
29. E
30. A
2
.
2
2. C. By definition, h  x   g 1  x  . Recalling that the domain of a function is equal to the range
of its inverse, h  1 means finding the value of x such that
2
2x  3
 1 , which is x 
x 1
3
3. B. Draw a right triangle such that the hypotenuse is of length x2  1 , the vertical leg is of
length x , and the horizontal leg is of length 1 , and let  be the angle opposite x . Then surely,
1
. Now we draw another right triangle such that the
  tan 1  x  , so cos    2
x 1
hypotenuse is again of length
tan   
x2  1 , but this time, the two legs are switched. Now,
1
.
x
4. C. This factors into y 
 x  3 x  2  x  2  . So, the asymptotes are
2
2
 x  2  x  2  x2  4
x  2, x  2, y  0 ,
there are 3 .
5. A. From Triangle Inequality, 2  AC  16 , so 3  4  5  ...  14  15  117 .
6. E. Squaring both sides and rearranging, we obtain x 2  y 2  25 , which we recognize as the
equation of a circle with radius 5 . Since this circle is symmetric around all four quadrants, the
area is
25
.
4
7. D. Let x, y, and z be the roots. What we need to find is
simplification yields
1 1 1
  , which, through
x y z
xy  xz  yz
, which we recognize as the quotient of the sum of the roots
xyz
3
taken two at a time, and the product of the roots. By Vieta’s formulas, this is
2
 3.
 1  1 2
3
8. A. A one-to-one function is a function that passes a horizontal line test. Hence, y  3 x  1 is
the only one-to-one function.
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Alpha Gemini Solutions
Mu Alpha Theta National Convention 2012
9. B. A rose of the form r  A sin  k  or r  A cos  k  has 2k petals if k is even, and k petals if
k is odd. Hence, this rose has 4024 petals.
10. D. The amplitude of f  x   A sin  x   B cos  x   C is
A2  B 2 . Hence,
112  602  61 .
11. A. The two digit palindromes are simply the two digit multiples of 11 , so there are 9 . Next
consider the form of three digit palindrome. There are nine numbers to choose for the first
digit (we cannot choose 0 ). Then the last digit has to be the same as the first, so there is only
one option for it. Finally, there are ten ways to select the middle digit. Hence, there are
9 10 1  90 three digit palindromes. Applying the same technique yields that there are
90 four digit palindromes, 900 five digit palindromes, and 900 six digit palindromes. So,
9  90  90  900  900  1989 .
12. D. First, prime factorize 12 as 2 2  3 . So, in order to make a number end in 0 in base 12 , we
need 4' s and 3's .We will count the number of both, and then take the smaller such number.
 2012   2012   2012   2012   2012   2012 
For 3, this is 





 1001
 3   9   27   81   243   729 
Next we count the 4’s. To do this, we count the 2’s, then divide by 2. Doing the above
method for 2, we obtain 2004, and hence there are 1002 factors of 4. Thus, it will end in
1001 zeros.
v v v v
v v v v
13. C. We know that u  v  u v sin  and u  v  u v cos  . So, by Pythagorean Theorem, the
hypotenuse is
v v2 v v 2
u  v  u  v  
v2 v2
v2 v2
v v
u v sin 2   u v cos 2   u v .
14. B. For a conic of the form Ax 2  Bxy  Cy 2  Dx  Ey  F  0 , the angle needed to eliminate
the xy-term is given by tan  2  
B
2
 2 .
. So, tan  2  
AC
45
15. B. Let y  e2 x . Then this equation translates to 2 y 2  3 y  1  0 . Factoring, this becomes
 2 y  1 y  1  0  y 
1
,1 . So, e 2 x  1  x  0 , and
2
  1  12 
 2
 2
1
1 1
e   x  ln    ln      ln 
 . So, the sum of the solutions is ln 
 .
 2  
2
2 2
2 
2 




2x
16. B. Converting to polar will be of more help. For a circle, we have x 2  y 2  r 2 , x  r cos   ,
and y  r sin   . We wish to maximize xy  r 2 sin   cos   . The maximum will occur
when sin    cos   , which happens when each is
18 
2 2

 9.
2 2
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2
. So, the maximum is
2
Alpha Gemini Solutions
Mu Alpha Theta National Convention 2012
17. D. Using Remainder Theorem, we evaluate f  1 . Doing so results in
f  1  1  2  3  4  ...  2000   1. Hence, the remainder is
2000  2001
 1  2001001 .
2
18. C. By properties of absolute value, we can rewrite this as 2 x  3  5 y  4  40 . Now, let us
shift our axes so that this is centered at the origin, and symmetric around all four quadrants.
Let u  x  3 and v  y  3 . Then this becomes 2 u  5 v  40 . Let us consider just the first
u v
  1 . Geometrically, this is a right triangle
20 8
8  20
 80 . Since this region is the same in all four
with side lengths 8 and 20 , so the area is
2
quadrant area. In the first quadrant, we have
quadrants, the total area is 4  80  320 .
19. D. Let z  cos 1  i sin 1 , since by de Moivre’s Theorem, z n  cos  n   i sin  n  . Then
n

1
z
consider the sum    
1 z
n 0  3 

3
3
. To rationalize this, multiply the
3  cos 1  i sin 1
numerator and denominator by 3  cos 1  i sin 1 to obtain
9  cos 1  9i sin 1
. So, our
10  6 cos 1
original sum, then, is simply the imaginary component of this sum, or
9sin 1
.
10  6 cos 1
20. C. A matrix is singular if its determinant is 0. Computing the determinant, this is
sin 2    cos2     cos  2  . If  cos  2   0 on the interval 0, 2  , then

 3 5 7
4
,
4
,
4
,
4
, for a total of 4 .
21. B. First, we put the ellipse into its standard form
 x  1
3
2
 y  2

2
2
 1 . Eccentricity is given by
a 2  b2
1
3


.
a
3
3
22. A. By half-angle formulas, cos 2  22.5  
means that
1  cos  45  2  2

. For a to be rational, this
2
4
2 2
2 2 2 2

 1.
is also a root. So the sum of the roots,    a   a 
4
4
4
 2  2  2  2  1
1 1
1

 .

The product of the roots, b  
.
The
product,
then,
is




8 8
 4  4  8
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Alpha Gemini Solutions
Mu Alpha Theta National Convention 2012
23. A. The perpendicular bisector passes through the midpoint, and has slope that is the negative
reciprocal of that of the original line. The midpoint is  1,5 , and the slope of the original
1
line  , so the slope of the perpendicular bisector is 3 . Applying point-slope form,
3
y  5  3  x  1  y  3x  8 . Using polar substitutions,
r  sin    3cos     8  r 
8
.
sin    3cos  
24. A. Let a be the term we are search for. Then x a   x 2 
6
3
4
   2 x    2   2160 .
 x 
 4
6 a
 x0  a  4 . So,
2
25. D. This can be rewritten as y 
sin  4 x  3 x   sin  4 x  3 x 
. Expanding using the sum and
cos  4 x  3 x   cos  4 x  3 x 
difference formulas, we obtain
sin  4 x  cos  3 x   sin  3 x  cos  4 x    sin  4 x  cos  3 x   sin  3 x  cos  4 x  
.
y
cos  4 x  cos  3x   sin  4 x  sin  3 x    cos  4 x  cos  3 x   sin  4 x  sin  3 x  
Cancelling, we have y 
2sin  4 x  cos  3 x  sin  4 x 


 tan  4 x  . Hence, the period is
.
2 cos  4 x  cos  3 x  cos  4 x 
4
26. E. These two lines have different slopes, so they will intersect somewhere. Hence, the
shortest distance between the two lines is 0 .
27. C. From the first equation, we have cos   
sin   
x3
, and from the second equation, we have
5
y 8
. Squaring both sides of each equation and adding them together, we have
9
 x  3
2
 y  8
2
. We recognize this as an ellipse, whose area is   5  9  45 .
25
81
28. A. The argument of a logarithmic function must be greater than 0 , so 2 x 2  x  3  0 .
3
Factoring, we obtain  2 x  3 x  1  0 . So, x   , and x  1 . Further, we cannot have a
2
1

0 in the denominator, which would occur at ln 1 . Setting the argument equal to 1 , we
obtain 2 x 2  x  3  1  2 x 2  x  4  0 . Applying quadratic formula, we have x 
1  33
.
4

 
 
 

Hence, the domain is  , 1  33    1  33 ,  3   1, 1  33    1  33 ,   .








4
4
2 
4
4

 
 

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Alpha Gemini Solutions
Mu Alpha Theta National Convention 2012
29. E. If we factor the argument of the product, we find that  x  7  is a root. This means that
f  7   0 , so the entire product will be 0 .
30. A. For this problem, we shall use a geometric approach. Let us make a graph plotting Goku’s
and Vegeta’s arrival times. We shall draw a square of length 120 (2 hours is 120 minutes),
and we’ll let Goku’s arrival time be on the x-axis, and Vegeta’s arrival time be on the y-axis.
The goal is to graph the successful region. We shall take a few examples. If Goku arrives at
12:00, then as long as Vegeta arrives between 12:00 and 12:20, they will have lunch. If Goku
arrives at 12:10, then as long as Vegeta arrives between 12:05 and 12:30, they will have
lunch. If Goku arrives at 12:15, then as long as Vegeta arrives between 12:10 and 12:35, they
will have lunch. So, we see the pattern. The successful region is the area bounded by
x  5  y  x  20 . This forms the stripe above. So then the probability is given by the area of
the stripe divided by the area of the square. Notice that the stripe splits the square into two
isosceles triangles of leg lengths 115 and 100 . Then, the probability is
1
1
2
2
1202  100   115 
223
2
2

2
120
1152
2:00
115
1:40
1:00
100
12:05
12:20
1:00
Goku
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1:55
2:00